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Which means that the blue side is going to have to look something like that. Just the way we drew it over here. This side down here, I'll make it a green side. This green side down here, we know nothing about. We never said that this is congruent to anything. If we knew, then we could use side-side-side. We only know that this side is congruent, this side is congruent, and this angle is congruent.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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This green side down here, we know nothing about. We never said that this is congruent to anything. If we knew, then we could use side-side-side. We only know that this side is congruent, this side is congruent, and this angle is congruent. So this green side, and I'll draw it as a dotted line, it could be of any length. We don't know what the length is of that green side. Now we have this magenta side.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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We only know that this side is congruent, this side is congruent, and this angle is congruent. So this green side, and I'll draw it as a dotted line, it could be of any length. We don't know what the length is of that green side. Now we have this magenta side. We have another side that's congruent here. This thing could pivot over here. We know nothing about this angle.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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Now we have this magenta side. We have another side that's congruent here. This thing could pivot over here. We know nothing about this angle. It could form any angle, but it does have to get to this other side. One possibility is that maybe the triangles are congruent. Maybe this side does go down just like that.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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We know nothing about this angle. It could form any angle, but it does have to get to this other side. One possibility is that maybe the triangles are congruent. Maybe this side does go down just like that. In which case, we actually would have congruent triangles. The kind of aha moment here, or the reason why SSA isn't possible, is that this side could also come down like this. There's two ways to get down to this base, if you want to call it that way.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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Maybe this side does go down just like that. In which case, we actually would have congruent triangles. The kind of aha moment here, or the reason why SSA isn't possible, is that this side could also come down like this. There's two ways to get down to this base, if you want to call it that way. You can come out that way, or you can kind of come in this way. That's why SSA by itself, with no other information, is ambiguous. It does not give you enough information to say that those triangles are definitely the same.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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There's two ways to get down to this base, if you want to call it that way. You can come out that way, or you can kind of come in this way. That's why SSA by itself, with no other information, is ambiguous. It does not give you enough information to say that those triangles are definitely the same. Now, there are special cases. In this situation, right over here, our angle, the angle in our SSA, our angle was acute. This is an acute angle right over here.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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It does not give you enough information to say that those triangles are definitely the same. Now, there are special cases. In this situation, right over here, our angle, the angle in our SSA, our angle was acute. This is an acute angle right over here. When you have an acute angle as one of the sides of your triangle, the other sides of the triangle, you could still have an obtuse angle. Remember, acute means less than 90 degrees. Obtuse means greater than 90 degrees.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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This is an acute angle right over here. When you have an acute angle as one of the sides of your triangle, the other sides of the triangle, you could still have an obtuse angle. Remember, acute means less than 90 degrees. Obtuse means greater than 90 degrees. You could still have an obtuse angle. That's why this is an option. One option is that you have two other acute angles.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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Obtuse means greater than 90 degrees. You could still have an obtuse angle. That's why this is an option. One option is that you have two other acute angles. This could be acute. This is also acute. Also acute.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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One option is that you have two other acute angles. This could be acute. This is also acute. Also acute. Also acute. You had the option where this is even more acute, even narrower, and then this becomes an obtuse angle. That is an obtuse angle.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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Also acute. Also acute. You had the option where this is even more acute, even narrower, and then this becomes an obtuse angle. That is an obtuse angle. That's only possible if you don't have two obtuse angles in the same triangle. You can't have two things that have larger than 90 degree measure in the same triangle. That's why there is a possibility where if you have another triangle that looks like this, and if I were to tell you very clearly that this angle right over here is obtuse, and that is the A in the SSA, I would say I have another triangle where this angle is congruent to that other triangle, some angle of that other triangle, and then one of the sides adjacent to it is congruent, and then the next side over is also congruent, but that's not so ambiguous because we could try to draw that.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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That is an obtuse angle. That's only possible if you don't have two obtuse angles in the same triangle. You can't have two things that have larger than 90 degree measure in the same triangle. That's why there is a possibility where if you have another triangle that looks like this, and if I were to tell you very clearly that this angle right over here is obtuse, and that is the A in the SSA, I would say I have another triangle where this angle is congruent to that other triangle, some angle of that other triangle, and then one of the sides adjacent to it is congruent, and then the next side over is also congruent, but that's not so ambiguous because we could try to draw that. Let's draw that same congruent obtuse angle. We know nothing about this side down here because we haven't said that that's necessarily congruent, so that could be of any length. We do know that this triangle is going to have the same length for this side of the angle.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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That's why there is a possibility where if you have another triangle that looks like this, and if I were to tell you very clearly that this angle right over here is obtuse, and that is the A in the SSA, I would say I have another triangle where this angle is congruent to that other triangle, some angle of that other triangle, and then one of the sides adjacent to it is congruent, and then the next side over is also congruent, but that's not so ambiguous because we could try to draw that. Let's draw that same congruent obtuse angle. We know nothing about this side down here because we haven't said that that's necessarily congruent, so that could be of any length. We do know that this triangle is going to have the same length for this side of the angle. It looks like this. Then we know that this side is also going to be the same length. We haven't told you anything about this angle right over here, so this side could pivot over here.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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We do know that this triangle is going to have the same length for this side of the angle. It looks like this. Then we know that this side is also going to be the same length. We haven't told you anything about this angle right over here, so this side could pivot over here. We could kind of rotate it over there, but there's only one way now that this orange side can reach this green side. Now the only way is this way over here. We were more constrained, or this case isn't ambiguous, because we used up our obtuse angle here.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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We haven't told you anything about this angle right over here, so this side could pivot over here. We could kind of rotate it over there, but there's only one way now that this orange side can reach this green side. Now the only way is this way over here. We were more constrained, or this case isn't ambiguous, because we used up our obtuse angle here. The A here is an obtuse one, so then it constrains what the triangle can become. I don't want to make you say, maybe SSA in general, if SSA you do not want to use it as a postulate. I just wanted to make it clear that there is this special case where if you know that the A in the SSA is obtuse, then it becomes a little bit less ambiguous.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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We were more constrained, or this case isn't ambiguous, because we used up our obtuse angle here. The A here is an obtuse one, so then it constrains what the triangle can become. I don't want to make you say, maybe SSA in general, if SSA you do not want to use it as a postulate. I just wanted to make it clear that there is this special case where if you know that the A in the SSA is obtuse, then it becomes a little bit less ambiguous. Then finally there's a circumstance, so this is an acute angle where it would be ambiguous. You have the obtuse angle, and then you have something in between, which is the right angle, where you have the A in SSA is a right angle. If you had it like this, if you have a right angle, and you have some base of unknown length, but you fix this length right over here, if you know that this is fixed, because you're saying it's congruent to some other triangle, and if you know that the next length is fixed, and if you think about it, this next side is going to be the side opposite the right angle, it's going to have to be the hypotenuse of the right angle, if you know that the only way you can construct this, and similar to the obtuse case, and if you know the length of this, the only way you can do it is to bring it down over here.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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I just wanted to make it clear that there is this special case where if you know that the A in the SSA is obtuse, then it becomes a little bit less ambiguous. Then finally there's a circumstance, so this is an acute angle where it would be ambiguous. You have the obtuse angle, and then you have something in between, which is the right angle, where you have the A in SSA is a right angle. If you had it like this, if you have a right angle, and you have some base of unknown length, but you fix this length right over here, if you know that this is fixed, because you're saying it's congruent to some other triangle, and if you know that the next length is fixed, and if you think about it, this next side is going to be the side opposite the right angle, it's going to have to be the hypotenuse of the right angle, if you know that the only way you can construct this, and similar to the obtuse case, and if you know the length of this, the only way you can do it is to bring it down over here. That actually does lead to another postulate called the right angle side hypotenuse postulate, which is really just a special case of SSA where the angle is actually a right angle. Here they wrote the angle first. You can view this as angle, side, side, and they were able to do it because now they can write right angle, and so it doesn't form that embarrassing acronym.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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If you had it like this, if you have a right angle, and you have some base of unknown length, but you fix this length right over here, if you know that this is fixed, because you're saying it's congruent to some other triangle, and if you know that the next length is fixed, and if you think about it, this next side is going to be the side opposite the right angle, it's going to have to be the hypotenuse of the right angle, if you know that the only way you can construct this, and similar to the obtuse case, and if you know the length of this, the only way you can do it is to bring it down over here. That actually does lead to another postulate called the right angle side hypotenuse postulate, which is really just a special case of SSA where the angle is actually a right angle. Here they wrote the angle first. You can view this as angle, side, side, and they were able to do it because now they can write right angle, and so it doesn't form that embarrassing acronym. This would also be a little bit common sense, because if you know two sides of a right triangle, and we haven't gone into depth in this in the geometry playlist, but you might already be familiar with it, by Pythagorean theorem, you can always figure out the third side. If you have this information about any triangle, you can always figure out the third side, and then you can use side, side, side. I just want to show you this little special case, but in general, the important thing is that you can't just use SSA unless you have more information.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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Because at least it looks that triangle CFE is similar to ABE. And the intuition there is it's kind of embedded inside of it. And we're going to prove that to ourselves. And it also looks like triangle CFB is going to be similar to triangle DEB. But once again, we're going to have to prove that to ourselves. And then maybe we can deal with all the ratios of the different sides to CF right over here. And then actually figure out what CF is going to be.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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And it also looks like triangle CFB is going to be similar to triangle DEB. But once again, we're going to have to prove that to ourselves. And then maybe we can deal with all the ratios of the different sides to CF right over here. And then actually figure out what CF is going to be. So first, let's prove to ourselves that these definitely are similar triangles. So you have this 90 degree angle in ABE. And you have this 90 degree angle in CFE.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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And then actually figure out what CF is going to be. So first, let's prove to ourselves that these definitely are similar triangles. So you have this 90 degree angle in ABE. And you have this 90 degree angle in CFE. If we can prove just one other angle or one other set of corresponding angles is congruent in both, then we've proved that they're similar. And we can either show that, look, they both share this angle right over here. Angle CEF is the same as angle AEB.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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And you have this 90 degree angle in CFE. If we can prove just one other angle or one other set of corresponding angles is congruent in both, then we've proved that they're similar. And we can either show that, look, they both share this angle right over here. Angle CEF is the same as angle AEB. So we've shown two angles, two corresponding angles in these triangles. This is an angle in both triangles. They are congruent, so the triangles are going to be similar.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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Angle CEF is the same as angle AEB. So we've shown two angles, two corresponding angles in these triangles. This is an angle in both triangles. They are congruent, so the triangles are going to be similar. You could also show that this line is parallel to this line, because obviously these two angles are the same. And so these angles will also be the same. So they're definitely similar triangles.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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They are congruent, so the triangles are going to be similar. You could also show that this line is parallel to this line, because obviously these two angles are the same. And so these angles will also be the same. So they're definitely similar triangles. So let's just write that down, get that out of the way. We know that triangle ABE is similar to triangle CFE. And you want to make sure you get in the right order.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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So they're definitely similar triangles. So let's just write that down, get that out of the way. We know that triangle ABE is similar to triangle CFE. And you want to make sure you get in the right order. F is where the 90 degree angle is. B is where the 90 degree angle is. And then E is where this orange angle is.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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And you want to make sure you get in the right order. F is where the 90 degree angle is. B is where the 90 degree angle is. And then E is where this orange angle is. So CFE, it's similar to triangle CFE. Now let's see if we can figure out that same statement going the other way, looking at triangle DEB. So once again, you have a 90 degree angle here.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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And then E is where this orange angle is. So CFE, it's similar to triangle CFE. Now let's see if we can figure out that same statement going the other way, looking at triangle DEB. So once again, you have a 90 degree angle here. If this is 90, then this is definitely going to be 90 as well. You have a 90 degree angle here at CFB. You have a 90 degree angle at DEF or DEB, however you want to call it.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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So once again, you have a 90 degree angle here. If this is 90, then this is definitely going to be 90 as well. You have a 90 degree angle here at CFB. You have a 90 degree angle at DEF or DEB, however you want to call it. So they have one set of corresponding angles that are congruent. And then you'll also see that they both share this angle right over here on the smaller triangle. So I'm now looking at this triangle right over here as opposed to the one on the right.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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You have a 90 degree angle at DEF or DEB, however you want to call it. So they have one set of corresponding angles that are congruent. And then you'll also see that they both share this angle right over here on the smaller triangle. So I'm now looking at this triangle right over here as opposed to the one on the right. So they both share this angle right over here, DBE. Angle DBE is the same as angle CBF. So I've shown you already that we have this angle is congruent to this angle.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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So I'm now looking at this triangle right over here as opposed to the one on the right. So they both share this angle right over here, DBE. Angle DBE is the same as angle CBF. So I've shown you already that we have this angle is congruent to this angle. And we have this angle as a part of both. So it's obviously congruent to itself. So we have two corresponding angles that are congruent to each other.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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So I've shown you already that we have this angle is congruent to this angle. And we have this angle as a part of both. So it's obviously congruent to itself. So we have two corresponding angles that are congruent to each other. So we know that this larger triangle over here is similar to this smaller triangle over there. So let me write this down. So we also know that triangle DEB is similar to triangle CFB.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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So we have two corresponding angles that are congruent to each other. So we know that this larger triangle over here is similar to this smaller triangle over there. So let me write this down. So we also know that triangle DEB is similar to triangle CFB. Now what can we do from here? Well, we know that the ratios of corresponding sides for each of those similar triangles are going to have to be the same. But we only have one side of one of the triangles.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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So we also know that triangle DEB is similar to triangle CFB. Now what can we do from here? Well, we know that the ratios of corresponding sides for each of those similar triangles are going to have to be the same. But we only have one side of one of the triangles. So in the case of AB and CFB, we've only been given one side. In the case of DEB and CFB, we've only been given one side right over here. So there doesn't seem to be a lot to work with.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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But we only have one side of one of the triangles. So in the case of AB and CFB, we've only been given one side. In the case of DEB and CFB, we've only been given one side right over here. So there doesn't seem to be a lot to work with. And this is why this is a slightly more challenging problem here. Let's just go ahead and see if we can assume one of the sides, or maybe a side that's shared by both of these larger triangles, and then maybe things will work out from there. So let's just assume that this length right over here, let's just assume that BE is equal to Y.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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So there doesn't seem to be a lot to work with. And this is why this is a slightly more challenging problem here. Let's just go ahead and see if we can assume one of the sides, or maybe a side that's shared by both of these larger triangles, and then maybe things will work out from there. So let's just assume that this length right over here, let's just assume that BE is equal to Y. So let me just write this down. This whole length is going to be equal to Y, because this at least gives us something to work with. And Y is shared by both ABE and DEB, so that seems useful.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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So let's just assume that this length right over here, let's just assume that BE is equal to Y. So let me just write this down. This whole length is going to be equal to Y, because this at least gives us something to work with. And Y is shared by both ABE and DEB, so that seems useful. And then we're going to have to think about the smaller triangles right over there. So maybe we'll call this length BFx. And then let's call FE.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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And Y is shared by both ABE and DEB, so that seems useful. And then we're going to have to think about the smaller triangles right over there. So maybe we'll call this length BFx. And then let's call FE. Well, if this is X, then this is Y minus X. So we've introduced a bunch of variables here, but maybe with all the proportionalities and things, just maybe things will work out, or at least we'll have a little bit more sense of where we can go with this actual problem. But now we can start dealing with the similar triangles.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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And then let's call FE. Well, if this is X, then this is Y minus X. So we've introduced a bunch of variables here, but maybe with all the proportionalities and things, just maybe things will work out, or at least we'll have a little bit more sense of where we can go with this actual problem. But now we can start dealing with the similar triangles. For example, so we want to figure out what CF is. We now know that for these two triangles right here, the ratio of the corresponding sides are going to be constant. So for example, the ratio between CF and 9, their corresponding sides, the ratio between CF and 9 has got to be equal to the ratio between Y minus X, that's that side right there, Y minus X and the corresponding side of the larger triangle.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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But now we can start dealing with the similar triangles. For example, so we want to figure out what CF is. We now know that for these two triangles right here, the ratio of the corresponding sides are going to be constant. So for example, the ratio between CF and 9, their corresponding sides, the ratio between CF and 9 has got to be equal to the ratio between Y minus X, that's that side right there, Y minus X and the corresponding side of the larger triangle. Well, the corresponding side of the larger triangle is this entire length. And that entire length right over there is Y. So it's equal to Y minus X over Y.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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So for example, the ratio between CF and 9, their corresponding sides, the ratio between CF and 9 has got to be equal to the ratio between Y minus X, that's that side right there, Y minus X and the corresponding side of the larger triangle. Well, the corresponding side of the larger triangle is this entire length. And that entire length right over there is Y. So it's equal to Y minus X over Y. So we could simplify this a little bit. Well, I'll hold off for a second. Let's see if we can do something similar with this thing on the right.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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So it's equal to Y minus X over Y. So we could simplify this a little bit. Well, I'll hold off for a second. Let's see if we can do something similar with this thing on the right. So once again, we have CF, its corresponding side on DEB. So now we're looking at the triangle CFB, not looking at triangle CFE anymore. So now when we're looking at this triangle, CF corresponds to DE.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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Let's see if we can do something similar with this thing on the right. So once again, we have CF, its corresponding side on DEB. So now we're looking at the triangle CFB, not looking at triangle CFE anymore. So now when we're looking at this triangle, CF corresponds to DE. So we have CF over DE is going to be equal to, so CF over DE is going to be equal to X is going to be equal to, let me do that in a different color, it's going to be equal to, I'm using all my colors, it's going to be equal to X over this entire base right over here, so this entire B, which once again we know is Y, so over Y. And now this looks interesting because it looks like we have three unknowns. We have CF, sorry, we know what DE is already.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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So now when we're looking at this triangle, CF corresponds to DE. So we have CF over DE is going to be equal to, so CF over DE is going to be equal to X is going to be equal to, let me do that in a different color, it's going to be equal to, I'm using all my colors, it's going to be equal to X over this entire base right over here, so this entire B, which once again we know is Y, so over Y. And now this looks interesting because it looks like we have three unknowns. We have CF, sorry, we know what DE is already. This is 12. I could have written CF over 12. The ratio between CF and 12 is going to be the ratio between X and Y.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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We have CF, sorry, we know what DE is already. This is 12. I could have written CF over 12. The ratio between CF and 12 is going to be the ratio between X and Y. So we have three unknowns and only two equations, so it seems hard to solve it first because there's one unknown, another unknown, another unknown, another unknown, another unknown, and another unknown. But it looks like I can write this right here, this expression, in terms of X over Y, and then we could do a substitution. So that's why this was a little tricky.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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The ratio between CF and 12 is going to be the ratio between X and Y. So we have three unknowns and only two equations, so it seems hard to solve it first because there's one unknown, another unknown, another unknown, another unknown, another unknown, and another unknown. But it looks like I can write this right here, this expression, in terms of X over Y, and then we could do a substitution. So that's why this was a little tricky. So this one right here we can rewrite as CF, let me do it in that same green color, we can rewrite it as CF over 9 is equal to Y minus X over Y, the same thing as Y over Y minus X over Y, over, or 1 minus X over Y. All I did is I essentially, I guess you could say, distributed the 1 over Y times both of these terms. Y over Y minus X over Y, or 1 minus X minus Y.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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So that's why this was a little tricky. So this one right here we can rewrite as CF, let me do it in that same green color, we can rewrite it as CF over 9 is equal to Y minus X over Y, the same thing as Y over Y minus X over Y, over, or 1 minus X over Y. All I did is I essentially, I guess you could say, distributed the 1 over Y times both of these terms. Y over Y minus X over Y, or 1 minus X minus Y. And this is useful because we already know what X minus Y is equal to. Sorry, X over Y is equal to. We already know that X over Y is equal to CF over 12.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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Y over Y minus X over Y, or 1 minus X minus Y. And this is useful because we already know what X minus Y is equal to. Sorry, X over Y is equal to. We already know that X over Y is equal to CF over 12. So this right over here I can replace with this, CF over 12. So then we get, this is a home stretch here, CF, which is what we care about, CF over 9 is equal to 1 minus CF over 12. And now we have one equation with one unknown, and we should be able to solve this right over here.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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We already know that X over Y is equal to CF over 12. So this right over here I can replace with this, CF over 12. So then we get, this is a home stretch here, CF, which is what we care about, CF over 9 is equal to 1 minus CF over 12. And now we have one equation with one unknown, and we should be able to solve this right over here. So we could add CF over 12 to both sides. So you have CF over 9 plus CF over 12 is equal to 1. We just have to find a common denominator here, and I think 36 will do the trick.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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And now we have one equation with one unknown, and we should be able to solve this right over here. So we could add CF over 12 to both sides. So you have CF over 9 plus CF over 12 is equal to 1. We just have to find a common denominator here, and I think 36 will do the trick. So 9 times 4 is 36, so if you had to multiply 9 times 4, you have to multiply CF times 4. So you have 4 CF, 4 CF over 36 is the same thing as CF over 9, and then plus CF over 12 is the same thing as 3 CF over 36. And this is going to be equal to 1.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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We just have to find a common denominator here, and I think 36 will do the trick. So 9 times 4 is 36, so if you had to multiply 9 times 4, you have to multiply CF times 4. So you have 4 CF, 4 CF over 36 is the same thing as CF over 9, and then plus CF over 12 is the same thing as 3 CF over 36. And this is going to be equal to 1. And then we are left with 4 CF plus 3 CF is 7. CF over 36 is equal to 1. And then to solve for CF, we can multiply both sides times the reciprocal of 7 over 36.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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And this is going to be equal to 1. And then we are left with 4 CF plus 3 CF is 7. CF over 36 is equal to 1. And then to solve for CF, we can multiply both sides times the reciprocal of 7 over 36. So 36 over 7, multiply both sides times that, 36 over 7. This side things cancel out. And we are left with our final, we get our drum roll now.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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And then to solve for CF, we can multiply both sides times the reciprocal of 7 over 36. So 36 over 7, multiply both sides times that, 36 over 7. This side things cancel out. And we are left with our final, we get our drum roll now. CF is equal to CF, so all of this stuff cancels out. CF is equal to 1 times 36 over 7, or it's just 36 over 7. And this was a pretty cool problem, because what it shows you is if you have two things, let's see, this thing is some type of a pole or a stick or maybe the wall of a building or who knows what it is.
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Challenging similarity problem Similarity Geometry Khan Academy (2).mp3
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We're asked to translate and dilate the unit circle to map it onto each circle. So this is the unit circle right over here. It's centered at zero comma zero. It has a radius of one. That's why we call it a unit circle. And when they say translate, they say move it around. So that would be a translation of it.
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Proof all circles are similar Mathematics II High School Math Khan Academy.mp3
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It has a radius of one. That's why we call it a unit circle. And when they say translate, they say move it around. So that would be a translation of it. And then dilating it means making it larger. So dilating that unit circle would be doing, whoops, I just translated it, would be doing something like that. So we're gonna translate and dilate this unit circle to map it onto each circle.
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Proof all circles are similar Mathematics II High School Math Khan Academy.mp3
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So that would be a translation of it. And then dilating it means making it larger. So dilating that unit circle would be doing, whoops, I just translated it, would be doing something like that. So we're gonna translate and dilate this unit circle to map it onto each circle. So for example, I can translate it so that the center is translated to the center of that magenta circle. And then I can dilate it so that it has been mapped onto that larger magenta circle. I can do that, let me do it for a few more.
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Proof all circles are similar Mathematics II High School Math Khan Academy.mp3
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So we're gonna translate and dilate this unit circle to map it onto each circle. So for example, I can translate it so that the center is translated to the center of that magenta circle. And then I can dilate it so that it has been mapped onto that larger magenta circle. I can do that, let me do it for a few more. I'm not gonna do it for all of them. Just to give you the idea of what they're talking about. So now I'm translating the center of my, it's no longer a unit circle.
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Proof all circles are similar Mathematics II High School Math Khan Academy.mp3
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I can do that, let me do it for a few more. I'm not gonna do it for all of them. Just to give you the idea of what they're talking about. So now I'm translating the center of my, it's no longer a unit circle. I'm translating the center of my circle to the center of the purple circle. And now I'm gonna dilate it so it has the same radius. And notice, I can map it.
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Proof all circles are similar Mathematics II High School Math Khan Academy.mp3
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So now I'm translating the center of my, it's no longer a unit circle. I'm translating the center of my circle to the center of the purple circle. And now I'm gonna dilate it so it has the same radius. And notice, I can map it. And so if you can map one shape to another through translation and dilation, then the things are by definition, they are going to be similar. So this is really just an exercise in seeing that all circles are similar. If you just take any circle and you make it have the same center as another circle, then you can just scale it, you can just scale it up or down to match the circle that you moved it to the center of.
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Proof all circles are similar Mathematics II High School Math Khan Academy.mp3
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What I'd like to do in this video is use some geometric arguments to prove that the slopes of perpendicular lines are negative reciprocals of each other. And so just to start off, we have lines L and M, and we're going to assume that they are perpendicular, so they intersect at a right angle. We see that depicted right over here. And so I'm gonna now construct some other lines here to help us make our geometric argument. So let me draw a horizontal line that intersects at this point right over here. Let's call that point A. And so let me see if I can do that.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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And so I'm gonna now construct some other lines here to help us make our geometric argument. So let me draw a horizontal line that intersects at this point right over here. Let's call that point A. And so let me see if I can do that. There you go. So that's a horizontal line that intersects at A. And now I'm gonna drop some verticals from that.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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And so let me see if I can do that. There you go. So that's a horizontal line that intersects at A. And now I'm gonna drop some verticals from that. So I'm gonna drop a vertical line right over here, and I'm gonna drop a vertical line right over here. And so that is 90 degrees, and that is 90 degrees, and I've constructed it that way. This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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And now I'm gonna drop some verticals from that. So I'm gonna drop a vertical line right over here, and I'm gonna drop a vertical line right over here. And so that is 90 degrees, and that is 90 degrees, and I've constructed it that way. This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles. And let me now set up some points. So that already said, that's point A. Let's call this point B.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles. And let me now set up some points. So that already said, that's point A. Let's call this point B. Let's call this point C. Let's call this point D. And let's call this point E right over here. Now, let's think about what the slope of line L is. So slope of, let me move this over a little bit.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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Let's call this point B. Let's call this point C. Let's call this point D. And let's call this point E right over here. Now, let's think about what the slope of line L is. So slope of, let me move this over a little bit. So slope, slope of L is going to be what? Well, that's, you could view line L as line, the line that connects point CA. So it's the slope of CA, you could say.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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So slope of, let me move this over a little bit. So slope, slope of L is going to be what? Well, that's, you could view line L as line, the line that connects point CA. So it's the slope of CA, you could say. This is the same thing as slope of line CA. L is line CA. And so to find the slope, that's change in Y over change in X.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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So it's the slope of CA, you could say. This is the same thing as slope of line CA. L is line CA. And so to find the slope, that's change in Y over change in X. So our change in Y is going to be CB. So it's gonna be the length of segment CB. That is our change in Y.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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And so to find the slope, that's change in Y over change in X. So our change in Y is going to be CB. So it's gonna be the length of segment CB. That is our change in Y. So it is CB over our change in X, which is the length of segment BA, which is the length of segment BA right over here. So that is BA. Now, what is the slope of line M?
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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That is our change in Y. So it is CB over our change in X, which is the length of segment BA, which is the length of segment BA right over here. So that is BA. Now, what is the slope of line M? So slope, slope of M, and we could also say slope of, we could call line M line AE, line AE, like that. Well, if we're going to go between point A and point E, once again, it's just change in Y over change in X. Well, what's our change in Y going to be?
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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Now, what is the slope of line M? So slope, slope of M, and we could also say slope of, we could call line M line AE, line AE, like that. Well, if we're going to go between point A and point E, once again, it's just change in Y over change in X. Well, what's our change in Y going to be? Well, our change in Y, well, we're gonna go from this level down to this level as we go from A to E. We could have done it over here as well. We're gonna go from A to E. That is our change in Y. So we might be tempted to say, well, that's just going to be the length of segment DE.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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Well, what's our change in Y going to be? Well, our change in Y, well, we're gonna go from this level down to this level as we go from A to E. We could have done it over here as well. We're gonna go from A to E. That is our change in Y. So we might be tempted to say, well, that's just going to be the length of segment DE. But remember, our Y is decreasing. So we're gonna subtract that length as we go from this Y level to that Y level over there. And what is our change in X?
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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So we might be tempted to say, well, that's just going to be the length of segment DE. But remember, our Y is decreasing. So we're gonna subtract that length as we go from this Y level to that Y level over there. And what is our change in X? So our change in X, we're going to go, as we go from A to E, our change in X is going to be the length of segment AD. So AD. So our slope of M is going to be negative DE.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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And what is our change in X? So our change in X, we're going to go, as we go from A to E, our change in X is going to be the length of segment AD. So AD. So our slope of M is going to be negative DE. It's going to be the negative of this length because we're dropping by that much. That's our change in Y over segment A, over segment AD. So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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So our slope of M is going to be negative DE. It's going to be the negative of this length because we're dropping by that much. That's our change in Y over segment A, over segment AD. So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar. And then we're going to be able to show that these are the negative reciprocal of each other. So let's show that these two triangles are similar. So let's say that we have this angle right over here.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar. And then we're going to be able to show that these are the negative reciprocal of each other. So let's show that these two triangles are similar. So let's say that we have this angle right over here. And let's say that angle has measure X, just for kicks. And let's say that we have, let me do another color for, let's say we have this angle right over here. And let's say that the measure, that that has measure Y.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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So let's say that we have this angle right over here. And let's say that angle has measure X, just for kicks. And let's say that we have, let me do another color for, let's say we have this angle right over here. And let's say that the measure, that that has measure Y. Well we know X plus Y plus 90 is equal to 180 because together they are supplementary. So I could write, I could write that X plus 90, plus 90, plus Y, plus Y, is going to be equal to, is going to be equal to 180 degrees. If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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And let's say that the measure, that that has measure Y. Well we know X plus Y plus 90 is equal to 180 because together they are supplementary. So I could write, I could write that X plus 90, plus 90, plus Y, plus Y, is going to be equal to, is going to be equal to 180 degrees. If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees. Is going to be equal to 90 degrees. These are algebraically equivalent statements. So is equal to 90 degrees.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees. Is going to be equal to 90 degrees. These are algebraically equivalent statements. So is equal to 90 degrees. And how can we use this to fill out some of the other angles in these triangles? Well, let's see. X plus this angle down here has to be equal to 90 degrees.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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So is equal to 90 degrees. And how can we use this to fill out some of the other angles in these triangles? Well, let's see. X plus this angle down here has to be equal to 90 degrees. Or you could say X plus 90 plus what is going to be equal to 180. I'm looking at triangle CBA right over here. The interior angles of a triangle add up to 180.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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X plus this angle down here has to be equal to 90 degrees. Or you could say X plus 90 plus what is going to be equal to 180. I'm looking at triangle CBA right over here. The interior angles of a triangle add up to 180. So X plus 90 plus what is equal to 180? Well X plus 90 plus Y is equal to 180. We already established that.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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The interior angles of a triangle add up to 180. So X plus 90 plus what is equal to 180? Well X plus 90 plus Y is equal to 180. We already established that. Similarly over here. Y plus 90 plus what is going to be equal to 180? Well, same argument.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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We already established that. Similarly over here. Y plus 90 plus what is going to be equal to 180? Well, same argument. We already know Y plus 90 plus X is equal to 180. So Y plus 90 plus X is equal to, is equal to 180. And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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Well, same argument. We already know Y plus 90 plus X is equal to 180. So Y plus 90 plus X is equal to, is equal to 180. And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other. They both have an angle of X, they both have a measure X, they both have an angle of measure Y, and they're both right triangles. So just by angle, angle, angle, so we could say by angle, angle, angle, one of our similarity postulates, we know that triangle E, triangle EDA, EDA is similar to triangle, to triangle ABC, to triangle ABC. And so that tells us that the ratio of corresponding sides are going to be the same.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other. They both have an angle of X, they both have a measure X, they both have an angle of measure Y, and they're both right triangles. So just by angle, angle, angle, so we could say by angle, angle, angle, one of our similarity postulates, we know that triangle E, triangle EDA, EDA is similar to triangle, to triangle ABC, to triangle ABC. And so that tells us that the ratio of corresponding sides are going to be the same. And so for example, we know, let's find the ratio of corresponding sides. We know that the ratio of, let's say CB to BA, so let's write this down. We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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And so that tells us that the ratio of corresponding sides are going to be the same. And so for example, we know, let's find the ratio of corresponding sides. We know that the ratio of, let's say CB to BA, so let's write this down. We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same. So the ratio of CB over BA, over BA is going to be equal to, is going to be equal to, well the corresponding side to CB, it's the side opposite the X degree angle right over here. So the corresponding side to CB is side AD, so that's going to be equal to AD over, what's the corresponding side to BA? Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same. So the ratio of CB over BA, over BA is going to be equal to, is going to be equal to, well the corresponding side to CB, it's the side opposite the X degree angle right over here. So the corresponding side to CB is side AD, so that's going to be equal to AD over, what's the corresponding side to BA? Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE. And so this right over here, this right over here we saw from the beginning, this is the slope, this is the slope of L. So slope, slope of L. And how does this relate to the slope of M? Notice the slope of M is the negative reciprocal of this. You take the reciprocal, you're going to get DE over AD, and then you have to take this negative right over here.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE. And so this right over here, this right over here we saw from the beginning, this is the slope, this is the slope of L. So slope, slope of L. And how does this relate to the slope of M? Notice the slope of M is the negative reciprocal of this. You take the reciprocal, you're going to get DE over AD, and then you have to take this negative right over here. So we could write this as the negative reciprocal of slope of M. Negative reciprocal, reciprocal of M's, of M's slope. And there you have it. We've just shown that if we start with, if we assume these L and M are perpendicular, and we set up these similar triangles, and we were able to show that the slope of L is the negative reciprocal of the slope of M.
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Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3
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So this first question says, what is the arc measure in degrees of arc AC on circle P below? So this is point A, that is point C. And when they're talking about arc AC, since they only have two letters here, we can assume that it's going to be the minor arc. And when we talk about the minor arc, there's two potential arcs that connect point A and point C. There's the one here on the left, and then there's the one on, there is the one on the right. And since C isn't exactly, isn't exactly straight down from A, it's a little bit to the right, the shorter arc, the arc with a smaller length, or the minor arc is going to be this one that I'm depicting here right on the right. So what is this arc measure going to be? Well, the measure of this arc is going to be exactly the same thing as in degrees as the measure of the central angle that intercepts the arc. So that central angle, let me do it in a different color, I'll do it in this blue color.
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Finding arc measures Mathematics II High School Math Khan Academy.mp3
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And since C isn't exactly, isn't exactly straight down from A, it's a little bit to the right, the shorter arc, the arc with a smaller length, or the minor arc is going to be this one that I'm depicting here right on the right. So what is this arc measure going to be? Well, the measure of this arc is going to be exactly the same thing as in degrees as the measure of the central angle that intercepts the arc. So that central angle, let me do it in a different color, I'll do it in this blue color. That central angle is angle CPA, angle CPA, and the measure of that central angle is going to be 70 degrees plus 104 degrees. It's going to be this whole thing right over there. So it's going to be 174 degrees.
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Finding arc measures Mathematics II High School Math Khan Academy.mp3
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So that central angle, let me do it in a different color, I'll do it in this blue color. That central angle is angle CPA, angle CPA, and the measure of that central angle is going to be 70 degrees plus 104 degrees. It's going to be this whole thing right over there. So it's going to be 174 degrees. 174 degrees. That's the arc measure in degrees of arc AC. Let's keep doing these.
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Finding arc measures Mathematics II High School Math Khan Academy.mp3
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So it's going to be 174 degrees. 174 degrees. That's the arc measure in degrees of arc AC. Let's keep doing these. So let me do another one. So this next one asks us, in the figure below, in the figure below, segment AD, so this is point A, this is point D, so segment AD is this one right over here. Let me see if I can draw that.
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Finding arc measures Mathematics II High School Math Khan Academy.mp3
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Let's keep doing these. So let me do another one. So this next one asks us, in the figure below, in the figure below, segment AD, so this is point A, this is point D, so segment AD is this one right over here. Let me see if I can draw that. That's AD right over there. AD and CE are diameters of the circle. So let me draw CE.
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Finding arc measures Mathematics II High School Math Khan Academy.mp3
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Let me see if I can draw that. That's AD right over there. AD and CE are diameters of the circle. So let me draw CE. So CE is, we're going to connect points C and E. These are diameters. So let me, so they go straight. Whoops, I'm using the wrong tool.
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Finding arc measures Mathematics II High School Math Khan Academy.mp3
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So let me draw CE. So CE is, we're going to connect points C and E. These are diameters. So let me, so they go straight. Whoops, I'm using the wrong tool. So those are, somehow I should, all right. So those are, whoops, how did that happen? So let me, somehow my pen got really big.
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Finding arc measures Mathematics II High School Math Khan Academy.mp3
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Whoops, I'm using the wrong tool. So those are, somehow I should, all right. So those are, whoops, how did that happen? So let me, somehow my pen got really big. All right. That'll be almost there. Okay.
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Finding arc measures Mathematics II High School Math Khan Academy.mp3
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So let me, somehow my pen got really big. All right. That'll be almost there. Okay. So CE, there you go. So those are both diameters of the circle. P, what is the arc measure of AB, of arc AB in degrees?
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Finding arc measures Mathematics II High School Math Khan Academy.mp3
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Okay. So CE, there you go. So those are both diameters of the circle. P, what is the arc measure of AB, of arc AB in degrees? So arc AB, once again, there's two potential arcs that connect point A and B. There's the minor arc, and since this only has two letters, we'll assume it's the minor arc. It's going to be this one over here.
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Finding arc measures Mathematics II High School Math Khan Academy.mp3
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P, what is the arc measure of AB, of arc AB in degrees? So arc AB, once again, there's two potential arcs that connect point A and B. There's the minor arc, and since this only has two letters, we'll assume it's the minor arc. It's going to be this one over here. There's a major arc, but to denote the major arc, they would have said something like AEB or ADB or arc ACB to make us go this kind of the, this long way around. But this is arc AB. So we, in order to find the arc measure, we just really have to find the measure of the central angle.
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Finding arc measures Mathematics II High School Math Khan Academy.mp3
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It's going to be this one over here. There's a major arc, but to denote the major arc, they would have said something like AEB or ADB or arc ACB to make us go this kind of the, this long way around. But this is arc AB. So we, in order to find the arc measure, we just really have to find the measure of the central angle. This is a central angle that intercepts that arc. We could even say it defines that arc in some way. So how can we figure out this angle?
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Finding arc measures Mathematics II High School Math Khan Academy.mp3
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So we, in order to find the arc measure, we just really have to find the measure of the central angle. This is a central angle that intercepts that arc. We could even say it defines that arc in some way. So how can we figure out this angle? This one's a little bit trickier. Well, the key here is to realize that this 93-degree angle, it is vertical to this whole angle right over here. And we know from geometry, which we're still learning as we do this example problem, that vertical angles are going to have the same measure.
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Finding arc measures Mathematics II High School Math Khan Academy.mp3
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